I have a 120/240 3 phase, 4 wire (center tap) delta system (B stinger). My 3 phase load is 80 kva per phase (A, B, and C). My single phase load is 34 kva phase A and 41 kva phase C. What are the amps for each phase and how is this calculated? What size (amp rating) panel is needed?

This question was posted at Mike Holts site, and got alot of responses, but everyone there is saying the previous poster is wrong...

I thought you have 240kva load for the 80kva/phase, and then the 34kva and 41kva are figured at single phase 240V?

So how do you do it? I be confused...

Dnk...

[This message has been edited by Dnkldorf (edited 02-09-2006).]

[This message has been edited by Dnkldorf (edited 02-09-2006).]

Here's how I got it figured:

First the 240kva load would be 578a per phase.

Next I took the 34kva and the 41kva, that would be 34kva on a 240 single phase load, so I would have 141a additional on those perspective phases, leaving 7kva, on the 120v load.

So phase A is 578+141=719a
B is 578a
c is 578+141+58=777a

777a X 125%= 970A = 1000a Breaker for ocp


Shoot it down, and tell me where I am going wrong pls...


Dnk..

OK - here's my initial cut at this. Start with the 3Ø component - take 240,000 VA (80,000 VA per phase) divided by 240 * SqRt of 3, which comes out to around 578 amps.

Next I'd ignore the ØA load as it just counteracts the ØC load. The ØC load is 41,000 - divide that by 120V for another 342 amps (on the C leg).

All in all, 578 + 342 = 920 Amps <-- highest leg (which is ØC).

Taking 920 amps X 1.25 = 1,150 amps, I'd vote for a 1,200 amp OCP.

Tomorrow at work I'll plug the raw data into a panel schedule and see what it says.

Radar




[This message has been edited by Radar (edited 02-09-2006).]

Dnkldorf,
Isn't it getting too late for stuff like this? Look at the clock!! I was just going to shut the puter down and though I would take a peck and found this.
Now, back to shutting the puter down.
Good night!

Dave, it's morning now. rise and shine.....

Turn on the puter and get them brain cells oscillating..


Dnk...

I think I know where I went wrong...

The 34kva on A and C should be 68Kva load at 240. That would be 283A, not the 141A.

So..A phase would be 861A
B remains 578
c would be 919

So, the 919 x 125% would be a 1200A Ocp?

Does that sound better?

(Radar, you editted your post and I just saw it now)


Dnk...

[This message has been edited by Dnkldorf (edited 02-10-2006).]

316 on the nuetral?

Dnk..

Dnk - I think the only current on the neutral is the unbalanced portion of the load between ØC & ØA. That is, the amount of ØC load that is not offset by the ØA load. This is 7,000 VA of ØC load (41KVA - 34KVA) divided by 120V = 58 amps.

Radar

Sounds right to me. The 3-ph load has no neutral component, the center-tap connection just acting a ground reference so that the phases sit at 120/208/120. Thus the neutral current will be just the difference between the extra single-phase load on A and C.

My brain hurts.
The description of your problem is poor.

Do you have a single 80KVA three phase load or (3) 80KVA single phase loads
or a single 240kVA three phase load?

Is your single phase loading connected A-N 34kVA @120V and C-N 41kVA @120V or is it a 34kVA @240 connected A-C and 7kVA C-N at 120V or 68kVA @240V connected A-N-C and 7kVA C-N at 120V?

JBD,
>>>My brain hurts<<<

No pain, no gain....

It wasn't my description, it was originaly posted at Mike Holts site. I just figured we could throw it around and see what we come up with...

I don't think it matters if they are 120V loads or 240V loads on the A,C phases. I think that is what threw me off.

I used the sq root of A sqr'd formula to get the nuetral current.(don't know how to post it) I am not sure how Radar and Paul get their nuetral current answer though....

Now get them brain cells moving and let's see some answers...

Dnk...

[This message has been edited by Dnkldorf (edited 02-10-2006).]

Guys, Gerald Newton's form is up and running again and I think he need a little help getting more posters. His form got hit real hard with spam about a year or so ago and he got tired of fighting it and shut it down. It's been back up for a couple of months an has been getting very little activity. The main site is: http://electrician2.com/
The forum: http://electrician2.com/elwwwboard/wwwboard.html
Gerald is an Alaska guy.

Dnk - you know that in a household 120/240V service, the neutral carries only the unbalanced portion of the line current, the difference between L1 current and L2 current. In your example here, ignoring that fact that ØB even exists for the moment, the same thing is true. On ØC, 34KVA of the 41KVA load is balanced by the 34KVA on ØA, and the resulting current flows between ØC and ØA, not involving the neutral.

The unbalanced part of the load, 7KVA, is seen between ØC and the neutral. 7,000VA divided by 120V is approx equal to 58 amps. That is to say, the unbalanced part is 58 amps and flows between ØC and neutral.

The SINGLE PHASE component of the loads are sumarized thus (Kirchoff's Current Law):
ØA = 284 amps
Neutral = 58 amps
ØC = 342 amps (284 + 58)

I'm not sure I'm any good at describing this stuff, hope this helps.

Radar

Radar, I got you, but that is a split phase, 180 degree phase system. The above, would be a 3 phase 120 degree phase system. no?

In a three phase system(I'm gonna try)


N= Sq root of (A^+B^+C^)-(AB+AC+CB)

^=squared

Is this the wrong formula?

Did you get the same phase currents as I did?


Dnk..

Dnk,

You are spot on. In a two leg _single_ phase system, you need two equal currents on each leg to balance to zero on the neutral. In a three leg _three_ phase system, you need _three_ equal currents on each leg to balance to zero on the neutral (I am assuming similar power factor and no harmonics).

If you had 10A on phase A, 10A on phase B, and 0A on phase C, then you would see _10A_ on the neutral.

Oh, for future reference, '^' is often use to mean 'raised to the power', so squared would be '^2' and cubed would be '^3'.

-Jon

Dnk - I think that although we are talking about a 3 phase system overall, the 120/240V 1Ø 3W between L1 & L3 is, in essence, single phase power (disregarding L2 and the other 2 transformers). Consider that single phase resi power is frequently also derived from 2 legs of a 3 phase system.

Anyway, the overall voltage between L1 and L3 (ØA & ØC) is 240V Single Phase with a grounded neutral connection at the centerpoint. IMHO there is little (if any) difference theory wise between this and a resi system.

I see what you're saying, if you connected the 3 line wires to a multi-trace O-scope, each sine wave would be 120º apart from the others. However, if you just look at L1 to L3 on a scope, you'd see a single 240VRMS sine wave, as there is but a single potential between L1 & L3.

Not sure I'm making sense here, please pass the gin. It's been a long week.

Radar

Your right Radar, however....

On a residential "split phase" 240v system the phase voltages are 180 out of phase, and so are the nuetral currents. The nuetral currents, when they meet at a central point, cancel each other out, because they are 180 apart.

Now on a 3 phase system, if we use 2 of the phases, this is a "single phase" 240V circuit. These currents are 120 degrees apart, so when the meet at a central point, they do not completely cancel, like in a "split phase" system.


The nuetral calculations are different...
And to make matters worse, Redsy just threw a hole in my calculations with a new thought, maybe he'll chime in.....

It has been a long week, and I have a weekend of shoveling the white stuff ahead of me...

So, who has the answers to the above problem? It's killing me.....

Or are you guys keeping me in suspense, just to be cruel and give me more grey hair?


Dnk....



[This message has been edited by Dnkldorf (edited 02-10-2006).]

Ignoring the multiple ambiguities as pointed out by JBD, and taking the wording at face value, I'd say that the 80 kva per phase is simply 80 kva, 3-phase. Just like 50 amps per phase is simply 50 amps, 3-phase.
Therefore, the amperages would appear to be
80,000/(240x1.73) = 192 amps
Then, taking the single phase numbers at face value, I'd assume L-N connections because the problem specifies individual phases. This means that phase A would have an additional 283 amps, and phase C, an additional 341 amps.
So...

Phase A = 192 + 283 = 475 amps
Phase B = 192 amps
Phase C = 192 + 343 = 535 amps


Then again, according to some hard-to-follow explanation to a similar sounding problem in "Delmar's Standard Textbook of Electricity" these numbers are probably wrong.




[This message has been edited by Redsy (edited 02-10-2006).]

I'm alergic to that white stuff, makes me start to shake and shivver. Blimey, that's why I live herre I guess.

I'm having a hard time seeing any real difference between the 120V components of a 120/240 delta service and a resi service right next door that's taken off the same primary. In the resi you just have some missing components (the other 2 transformers). Kind of a new definition of a really really open delta.

Let's back up a little and change the original situation just a little. Supposing we had 34KVA on ØA and the same on ØC (instead of 41KVA). That would essentially equate to 68KVA across a 240V potential and render the neutral connection an anchor or ground reference point without any current. Current would flow in A and out C, or visa versa at any given instant of time, neutral current would be zero, A-C current would be 284 amps.

I'm still seeing a single 240V potential between 2 points (ØA & ØC) with a grounded center point, nothing more. You've got a single transformer secondary we're dealing with here, the other transformers connected to L2 (ØB) are of no consequence to the 120/240 part of this. IFF (that's if and only if) any of the 120V current also flowed thru the other transformers to L2 (B), then we'd have phase angles and some trig to work out, but that's not the case. All 120V current flow is limited to the one transformer secondary between L1 & L2. We're splitting that in half, so it appears to be a 180º split.

True enough the three phase currents are flowing all around the transformer connections following the 3 distinct sine waves, but the single phase 120/240 leg is not a split phase, it's just 240 volts.

It's getting late (even here), so maybe we'll just agree to disagree. But I will think about it some more.

Take care, my friend. Don't freeze out there tomorrow (eheh - take lots of warm up breaks). Sorry this got longer than I had intended.

Radar

Redsy - I read it that way (80KVA total 3Ø load) at first and had to erase my first post and refigure. It could easily be that, or you could have, say, an 80KW heater connected to each of the 3 pairs of terminals, which is how I've been looking at this.

The way I figured this is as follows:

<OL TYPE=1>

[*] 3 Phase Load = 192.70 Amps (80 KVA),


[*] 1 Phase "240 Volt" Load = 141.67 Amps (34 KVA),


[*] 1 Phase "120 Volt" Load = 58.34 Amps (7 KVA).
</OL>

Nominal Totals Per Line ("Phase"):
<OL TYPE=A>

[*] Line "A": 334.37 Amps (80 KVA @ 240v 3Ø + 34 KVA @ 240v),


[*] Line "B": 192.70 Amps (80 KVA @ 240v 3Ø),


[*] Line "C": 392.71 Amps (80 KVA @ 240v 3Ø + 34 KVA @ 240v + 7 KVA @ 120v).
</OL>

Panelboard + Feeder Size

Nominal Load Values WITH LCL:

* Panelboard: 600 Amps - 240/120V 3Ø 4 Wire
* Feeder: 1,000 MCM THHN/THWN Cu. (or equivalent Parallel Feeders, protected by 500 Amp 3 Pole OCPD).


Nominal Load Values WITHOUT LCL:

* Panelboard: 400 Amps - 240/120v 3Ø 4 Wire
* Feeder: 600 MCM THHN/THWN Cu. (or equivalent Parallel Feeders, protected by 400 Amp 3 Pole OCPD).

BTW: Load on the Grounded (Neutral) Conductor would be 7 KVA (58.34 Amps)

Scott35

I've never been involved with delta systems, so perhaps my figuring is way off base, but I'm with Radar.

The way I see it, although there is a basic 120-degree difference between phase currents, the A-C winding with its center-tap neutral is introducing another factor to the overall picture.

Whenever you have a tapped transformer winding, if you use that tap as your reference point, then the voltages at each end will be 180 degrees out of phase with each other.
Let's build up a system:

Start with just a single center-tapped winding, like a 3-wire 1120/240V service, and let's still call the hot legs A and C. We know that the neutral current will be the difference between A and C. Now add one extra winding from A to provide a B-phase in the form of an open delta, but don't connect any load to that phase. Obviously there can be no current in that winding, so the neutral current will remain unchanged.

Now add a third winding B-to-C to close the delta, but still leave no load on the B-phase. Will the neutral current change then? I don't think so, because the basic voltage/current relationships between the phases are still the same. There will be very little (in a theoretically perfect arrangement zero) current flowing in the A-B and B-C windings, will there not? So the A-C winding is still acting just like a 3-wire 120/240V service.

Now add a 3-ph delta-connected load to A-B-C. Obviously there will now be current flowing in all three windings, but will that affect the neutral?

I do not believe so, since the 3-ph load has no neutral connection. The current in the A-C winding will increase, but as the voltages on the A and C phases have not changed with respect to neutral/ground, the single-phase loads connected A-N and C-N are still drawing the same amount of current, and thus the neutral will still carry just the difference of the single-phase loads on A and C.

I will TRY (it is rather lengthy, and somewhat confusing, but then again, I'm dumb) to recount the explanation in the above-mentione book when I get home from work today, assuming I do not stop for beer & wings (wink wink Dink).
That is a big assumption, based on my recent workload and resulting inability to feed my beer & wing habit for several weeks!

Whoops, I missed that this was a _delta_ system with a center tap.

Dnk, The equation that you were using, and which I said was 'spot on', applies when you have a 120 degree phase angle to phases A,B, and C around the neutral. If the power factors deviate at all from being the same, so that the current flows are no longer 120 degrees out of phase, then the equation no longer applies.

Remember that voltage is always measured relative to some reference zero. If you pick a different reference point, then you will measure _different_ voltages. No matter which reference point you select, if you do all of the math and calculate the total work involved to move an electron from on point to another, the overall result will come out the same, but some of the numbers in the middle will look different.

Generally we use the grounded transformer terminal as our reference zero. But remember the discussions of corner grounded delta systems; _any_ transformer terminal could be grounded, and any transformer terminal could be our reference zero for calculations.

From the 'point of view' of the center tap in the delta system, phases A and C _are_ 180 degrees out of phase. If we created a virtual neutral (say by using a zig-zag transformer) and used that virtual neutral as our reference zero, then phases A and C would be 120 degrees out of phase. If we picked phase B as our reference zero, then phases A and C are only 60 degrees out of phase.

It probably makes the most sense to pick the center tap as the voltage reference zero. *grin*

I am going to assume that all of the loads are resistive, just to go through the calculation. This is almost certainly an incorrect assumption; there are probably motors in that three phase load. But if you know the power factors for the various loads, then you can adjust the phase angles for the individual current flows appropriately.

In my calculations below, I calculate the current flowing _into_ each terminal. This gives a current phase angle that is approximately 180 degrees out of phase with the voltage phase angle of that terminal. It is interesting to note that even with the assumption of resistive loads, the current flowing from each terminal is not quite in phase with the voltage at each terminal. An unbalanced three phase resistive load does not have unity power factor.

I am also assuming a total KVA of 315; 240KVA of 3 phase load plus 75KVA of single phase load. You will get different numbers if you read the question differently.

Finally, in my calculations below, the loads are very carefully double counted, because I count the current from phase C to phase A and again as the current from phase A to C. This is compensated by the phase angle calculations; as a check, take the sum of all phase currents times 240/root(3) and you will get roughly the same total VA as the sum of the original specifications.

I will select the voltage measured from phase A to the Grounded center tap as the phase reference zero.
This gives us the following voltage phase angles for the single phase loads:
A to G 0 degrees 120V
G to A 180 degrees 120V
C to G 180 degrees 120V
G to C 0 degrees 120V

For the three phase loads, we assume that 80 KVA is connected between each pair A-B, B-C, C-A, so we need the three phase angles:
A to C 0 degrees 240V
C to B 120 degrees 240V
B to A 240 degrees 240V
A to B 60 degrees 240V
B to C 300 degrees 240V
C to A 180 degrees 240V

First the neutral current:
Our A to G load is 34 KVA at 120V, so 283.33A at 0 degrees
Our C to G load is 41 KVA at 120V, so 341.67A at 180 degrees
283.33A at 0 plus 341.67A at 180 gives
58.33A at 180 degrees

Now the phase A current:
Our G to A current is 283.33A at 180 degrees (just the opposite of the A to G current)
Our B to A load is 80 KVA at 240V, so 333.33A at 240 degrees
Our C to A load is 333.33A at 180 degrees
Now we add up 283.33 at 180 degrees with 333.33 at 240 degrees and 333.33 at 180 gives
834.83A at 200 degrees

Phase B current:
Our G to B current is 0
Our A to B current is 333.33A at 60 degrees
Our C to B current is 333.33A at 120 degrees
gives 577.35A at 90 degrees

Phase C current
G to C is 341.67A at 0 degrees
A to C is 333.33A at 0 degrees
B to C is 333.33A at 300 degrees
gives 889.8A at 341 degrees

I'll leave selection of OCPD and conductor size for someone else

-Jon

If you look at the current flowing through the coil connected between A and C phases, you will see that this is strictly single phase. There can be no phase shift from one end of a transformer winding to the other end(even if there is a center-tap).

The normal 120/240 single phase calculations apply for any load connected A-N, N-C.
I=kVA/E, use L-L kVA if using L-L voltage

Normal three phase calculations can be used for any loads connected A-B-C.
I=kVA/(E*1.73), use L-L kVA if using L-L voltage

Now just add up the line currents.

JBD:

Not quite. Your statement is correct that you calculate the single phase currents as single phase currents and the three phase currents as three phase currents. However the phase angle of the phase A three phase currents will be different from the phase A single phase currents, so you need to use vector addition to 'just add them up'.

-Jon

The phase currents in windings AB and BC are

80KVA/240V = 333A.

These currents combine to yield,

Ib = 333Ax1.732 = 577A for the 80 kva per phase.
Single phase =34 + (34+ 7) kva = 68 + 7 kva.
For winding CA, 68KVA of the single phase load may be treated as a 240V load and added to the 80KVA load to yield 148KVA of 240V load . Phase (not line) current is 148KVA/240 = 616A. To obtain the line current, Ia, we combine vectorially the phase current from winding AB.

Ia = 616A + 333A[cos(60) +jsin(60)] = 783A + j289A yields 835A magnitude.

The remaining 7KVA of single phase load adds 58.3A into node C. Then

Ic = 616A + 58.3A + 333A[cos(60) +jsin(60)] = 841A + j289A yields 890A magnitude.

Ia = 835A
Ib = 577A
Ic = 890A

winnie,

You can not have a phase angle difference on a single conductor. Yes the current flowing in the C-B winding will be at a different phase angle from the current in the C-A winding, but there is still only 1 current in the C phase conductor.

Go ahead and work out the math for the summation of these two winding currents at the C conductor and then tell me if it is substantially different from my method. I don't think it is worth the effort for sizing a conductor and overcurrent protective device. Because if we want to get down and dirty exact, we really need to look at the power factor of each load before we try to perform vector addition.

More questions than answers are generated based on the ambiguously worded question--

However, remember that in a delta system, the phase current (L-N) is less than the line currents(L-L) by a factor of 1.73.
Therefore, in order to determine line currents, and the required OCPDs, wouldn't we need clarification of the terms "single phase load"?

Is there an engineer in the house?

"Is there an engineer in the house?"
I just worked it out for you. So did winnie.
Our answers are the same.

[This message has been edited by Bob (edited 02-11-2006).]

The "no pain, no gain" statement has come back to haunt me.......

I'm almost out of smokes, I'm getting snowed in, went through alot of scrap paper, and my fingers hurt from erasing so much, but I'm still thinking...

I hope to have these phase angles figured out by the end of this weekend, then I'll throw another example at you, if you guys don't mind, so I can double check myself...

Redsy, I may take you up on the book offer...


Dnk..

JBD,

I agree with you that there is only one phase angle for the current flowing in the wire.

To calculate the total current flow, you have to account for the phase angle differences in the summation.

I do agree with you that the result that you get by simply adding the single phase and three phase currents without considering phase angle will probably get you as close as calculating in the angles without knowing the power factor of the loads and thus the actual phase current angles.

-Jon

posted by JBD
"You can not have a phase angle difference on a single conductor. Yes the current flowing in the C-B winding will be at a different phase angle from the current in the C-A winding, but there is still only 1 current in the C phase conductor."

That is correct. All load on winding AC are in phase. They are not in phase with AB and BC.

"Go ahead and work out the math for the summation of these two winding currents at the C conductor and then tell me if it is substantially different from my method. I don't think it is worth the effort for sizing a conductor and overcurrent protective device."
If you add the phase load in AC to the single phase load you get 625 amps.
Just adding phase AC to AB you get 625 + 333
= 958 amps. Same for BC.
If load is contineous 958 x 1.25 = 1197.
Doing it with vectors you get
Ia = 835A
Ib = 577A
Ic = 890A
890 x 1.25 = 1112.
The higher the single phase load the greater the error.



[This message has been edited by Bob (edited 02-12-2006).]

Please, lets not get caught up too much in this calculation; remember that all these calculations are _always_ approximations. We assume sinusoidal loads. We assume nominal voltage. We have been assuming unity power factor. We assume lots of things.

Lets try things by 'ignoring' phase angles. First we calculate the current for the three phase load, using VA/ V / root(3)
A 'pure' 240kVa 240V three phase load is 577A/phase (approximately!) Of course, this calculation automatically includes the phase angle differences between the various three phase legs.

Then we calculate the single phase loads using VA/V
A 34 KVA 120V single phase load is 283A
A 41 KVA 120V single phase load is 342A

Then we just add the numbers up on the various legs. This, of course, neglects the phase angle difference between the single and three phase load, giving.
A: 860A B 577A C 919A

This as compared to the calculation which includes the phase angle differences:
A:835A B577A C 890A

(Bob and I appear to be getting exactly the same numbers using different calculation methods, which IMHO is a rather good check)

In other words, by ignoring the phase difference one gets about 3% difference in the calculated currents versus including the phase angles but assuming that the three phase load has a unity power factor.

Now, let us change our assumptions, and use a 0.8 lagging power factor for the 3 phase load. This is a lagging phase angle of about 37 degrees. If I put the new phase angles into the calculation, I get:
A:919A B 577A C 771A

That is to say that our initial assumption of unity power factor for the three phase load leads to a much larger potential error than just adding currents up and ignoring the phase difference between the single and the three phase loads.

-Jon

Jon, That's all I was trying to say. We are splitting hairs over how precise our inprecise guess is. Without acknowledging the assumptions that have been made.

I have no problem with people trying to learn the correct methods, but there is a time and a place for everything. I just felt this was the time to learn how to do " engineering based estimation" which is also a valuable tool.

[This message has been edited by JBD (edited 02-12-2006).]

So, whaddaheck is the correct answer???

Ironically, I have been performing A LOT of Load Calcs + Panel Schedules -vs- Actual In-Field Amperage Tests (before and after the Calcs + Schedules were done), and so far things have been very accurate.

A couple systems (4, to be exact) were 4 Wire "Closed" Delta (3 Transformers, middle one with center tap), and an additional two were 4 Wire "Open Delta Vee Connections" (2 Transformers, largest has center tap).

Had several 3 Wire Deltas (2 Separately Derived Systems, 3 from the Utility guys), ranging from 240V 3Ø 3 Wire Corner Grounded Deltas, to 600V 3Ø 3 Wire Ungrounded Deltas.

Most systems are 4 Wire Wye setups - either 208Y/120V or 480Y/277V.

Anyhow, what's the verdict?

Scott35

Verdict?

We don't need no steenkin' verdict.

I think that the best that we can do is cover the possible different assumptions, and play with different calculation techniques.

I was describing using vector addition of the various currents, but that requires making assumptions about the power factors of the loads.

As JBD notes, a rough and ready estimate simply using addition and ignoring phase angles will give us a number that falls right into the range of possible values that you get with the "more accurate" vector calculations as you change the power factor assumptions. IMHO a very useful reminder that sometimes you don't really care if 86 angels or 94 angels can dance on a particular pin, and "about 90" is the correct answer to use

-Jon